Rewrite the function by completing the square. $f(x)=x^{2}+8x-29$ $f(x)=(x+$
Explanation: We want to complete $x^2{+8}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+8}}{2}\right)^2={16}$ to it: $x^2{+8}x+{16}=(x+4)^2$ In order to keep the expression equivalent, we add and subtract ${16}$, not forgetting the expression's constant term, $-29$ : $\begin{aligned} f(x)&=x^2+8x-29 \\\\ &=x^2+8x+{16}-29-{16} \\\\ &=(x+4)^2-29-16 \\\\ &=(x+4)^2-45 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 4)^2 - 45$